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Zorluk: KolayResistance and Resistivity

A continuous spool of aluminum hookup wire has a total length of 80 meters80\text{ meters} and a measured electrical resistance of 2.4 Ω2.4\ \Omega. If an electronics technician cuts a 20-meter20\text{-meter} piece from this spool for a project, what is the electrical resistance of the cut piece in ohms (Ω\Omega)?

Cevap: 0.6 Ω

Cevap

0.6 Ω
The electrical resistance of a uniform wire is directly proportional to its length (R=ρLAR = \rho \frac{L}{A}). Since 20 m20\text{ m} is one-fourth (14\frac{1}{4} or 0.250.25) of the original 80 m80\text{ m} length, the resistance of the cut piece is one-fourth of the total resistance: 2.4 Ω×0.25=0.6 Ω2.4\ \Omega \times 0.25 = 0.6\ \Omega.

Adım Adım Çözüm

1
Identify the physical relationship governing wire resistance and length.
Resistance is directly proportional to length: R=ρLAR = \rho \frac{L}{A}, which means R2=R1×L2L1R_2 = R_1 \times \frac{L_2}{L_1}.
Because the material (resistivity ρ\rho) and cross-sectional area (AA) remain constant, resistance scales directly with length.
2
Find the proportion of the new length relative to the original length.
LcutLtotal=20 m80 m=0.25\frac{L_{\text{cut}}}{L_{\text{total}}} = \frac{20\text{ m}}{80\text{ m}} = 0.25
This establishes what fraction of the total opposition to electron flow the shorter segment represents.
3
Calculate the resistance of the 20-meter wire segment.
Rcut=2.4 Ω×0.25=0.6 ΩR_{\text{cut}} = 2.4\ \Omega \times 0.25 = 0.6\ \Omega
Multiplying the total resistance by the length ratio yields the exact resistance of the cut section.

Anahtar Kavram

The electrical resistance of a conductor is directly proportional to its length (R∝LR \propto L) when cross-sectional area and material remain constant.
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